# proving that a ring does not have ACCP

I am trying to prove the following: the ring $$R=\frac{\mathbb{Z}[x_1,x_2,x_3,\cdots]}{(x_1-x_2^2,x_2-x_3^2,\cdots)}$$does not have the ascending chain condition for principal ideals.

By construction, we have $$x_i=x_{i+1}^2$$ in $$R$$. So there is an obvious candidate for an infinite chain of principal ideals, namely, $$(x_1)\subset(x_2)\subset\cdots.$$ However, I have been unable to show that the inclusions in this chain (or at least infinitely many of them) are strict. I have managed to prove it (see below), but I am unsure if my argument is correct. So please tell me

• if my argument is valid, and
• if there is a better way to prove this.

Any help is appreciated. Thanks in advance!

My attempt: Consider the ring homomorphism $$\varphi:\mathbb{Z}[x_1,x_2,\cdots]\to\mathbb R$$ which extends the inclusion $$\mathbb{Z}\to\mathbb R$$ and maps $$x_i\to 2^{2^{-n}}$$. Because $$\left(2^{2^{-(n+1)}}\ \ \right)^2=2^{2^{-n}}$$, the kernel of this homomrphism contains the ideal $$(x_1-x_2^2,x_2-x_3^2,\cdots)$$. So it descends to a surjective ring homomorphism $$\overline{\varphi}:R\to\operatorname{im}\varphi$$.

Now suppose that $$(x_1)=(x_2)$$ in $$R$$. Since $$\overline{\varphi}((x_n))=(2^{2^{-n}})$$, this means that $$(2^{2^{-1}})=(2^{2^{-2}})$$ in $$\operatorname{im}\varphi$$. But $$\operatorname{im}\varphi$$ is an integral domain and its units are $$\pm 1$$, so $$2^{2^{-1}}=\pm 2^{2^{-2}}$$, a contradiction.

## 1 Answer

There's a big gap in your proof: you've never shown that the units of $$\operatorname{im}\varphi$$ are just $$\pm 1$$. In fact, this is false: for instance, $$(\sqrt{2}+1)(\sqrt{2}-1)=1$$ so $$\sqrt{2}\pm 1$$ are units in $$\operatorname{im}\varphi$$.

Here's how I would show it. Suppose one of the inclusions is not strict, i.e. that $$x_i$$ divides $$x_{i+1}$$ for some $$i$$. The fact that $$x_i$$ divides $$x_{i+1}$$ is witnessed by finitely many of the generators and relations of $$R$$: there is a polynomial $$p\in \mathbb{Z}[x_1,x_2,\dots]$$ such that $$x_ip-x_{i+1}\in (x_1-x_2^2,x_2-x_3^2)\dots$$, and then there are only finitely many variables appearing in $$p$$ and finitely many variables needed to write $$x_ip-x_{i+1}$$ as a linear combination of $$x_1-x_2^2,x_2-x_3^2,\dots$$. This means that there is some $$n$$ such that $$x_i$$ divides $$x_{i+1}$$ in the ring $$R_n=\mathbb{Z}[x_1,\dots,x_n]/(x_1-x_2^2,x_2-x_3^3\dots,x_{n-1}-x_n^2)$$ which has only the generators and relations of $$R$$ up to $$x_n$$. But $$R_n$$ is easily seen to be isomorphic to just $$\mathbb{Z}[x]$$, by mapping $$x_j$$ to $$x^{2^{n-j}}$$. So to say that $$x_i$$ divides $$x_{i+1}$$ would mean that $$x^{2^{n-i}}$$ divides $$x^{2^{n-i-1}}$$ in $$\mathbb{Z}[x]$$ which is false.

Alternatively, for an approach that is closer to yours, you could define a ring $$\mathbb{Z}[x,x^{1/2},x^{1/4},\dots]$$ of formal linear combinations of nonnegative dyadic rational powers of $$x$$ (i.e., the monoid ring of the monoid of nonnegative dyadic rational numbers). You can then define a homomorphism $$\varphi:R\to\mathbb{Z}[x,x^{1/2},x^{1/4},\dots]$$ sending $$x_n$$ to $$x^{2^{-n}}$$ (in fact, this $$\varphi$$ is an isomorphism). If $$x_i$$ divided $$x_{i+1}$$, we would conclude that $$x^{2^{-i}}$$ divides $$x^{2^{-i-1}}$$ and so they differ by a unit. But the only units of $$\mathbb{Z}[x,x^{1/2},x^{1/4},\dots]$$ are $$\pm 1$$ (because degrees add when you multiply, just like with ordinary polynomials), and so this is a contradiction.